(a) Explanation Definition used: Definition 1: “The equation of the shifted hyperbola with center as ( h , k ) and horizontal axis ( x -axis) axis as transverse axis is ( x − h ) 2 a 2 − ( y − k ) 2 b 2 = 1 , when a > 0 , b > 0 , c 2 = a 2 + b 2 ”. Definition 2: “The equation of the hyperbola with center at the origin, vertices ( − a , 0 ) , ( a , 0 ) and foci F 1 ( − c , 0 ) , F 2 ( c , 0 ) is x 2 a 2 − y 2 b 2 = 1 , where c 2 = a 2 + b 2 , a > 0 , b > 0 ”. Definition 3: “The equation of the asymptotes of the hyperbola ( x − h ) 2 a 2 − ( y − k ) 2 b 2 = 1 are ( x − h ) a + ( y − k ) b = 0 and ( x − h ) a − ( y − k ) b = 0 ”. Definition 4: “If h and k are any two positive real numbers and a graph of any equation is in terms of x and y then. ( i ) Replacing x by x − h , shift the graph right to h units. ( i i ) Replacing x by x + h , shift the graph left to h units. ( i i i ) Replacing y by y − k , shift the graph upward to k units. ( i v ) Replacing y by y + k , shift the graph downward to k units.” Calculation: The given equation of conic is ( y + 2 ) 2 36 − x 2 64 = 1 (1) Now, by the definition 1, the equation (1) is a shifted hyperbola y 2 36 − x 2 64 = 1 (2) That is, equation ( 1 ) may obtain from equation ( 2 ) replacing x by x and y by y + 2 . Therefore, first find the center, vertices and foci of the hyperbola ( 2 ) and then keep the x -coordinates remains same and y -coordinates by 2 units downward. Compare the equation (2) with the hyperbola y 2 a 2 − x 2 b 2 = 1 . a 2 = 36 a = 6 b 2 = 64 b = 8 Clearly, a = 6 , b = 8 and a > 0 , b > 0 . By definition 2, c 2 = a 2 + b 2 c 2 = 36 + 64 c 2 = 100 c = 10 Now, find the center, vertices and foci of the unshifted hyperbola represented by equation (2) and then obtain the respective values for the shifted hyperbola represented by equation (1) as follows in Table 1 (b) To determine To sketch: The graph of the hyperbola ( y + 2 ) 2 36 − x 2 64 = 1 .

Precalculus: Mathematics for Calculus (Standalone Book)

7th Edition

ISBN: 9781305071759

Author: James Stewart, Lothar Redlin, Saleem Watson

Publisher: Brooks Cole

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Chapter 11.4, Problem 26E

(a)

To determine

To find: The center, vertices, foci and asymptotes of the hyperbola (y+2)236−x264=1 .

(b)

To determine

To sketch: The graph of the hyperbola (y+2)236−x264=1 .

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Chapter 11.4, Problem 25E

Chapter 11.4, Problem 27E

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Chapter 11 Solutions

Precalculus: Mathematics for Calculus (Standalone Book)

Ch. 11.1 - A parabola is the set of all points in the plane...Ch. 11.1 - The graph of the equation x2 = 4py is a parabola...Ch. 11.1 - The graph of the equation y2 = 4px is a parabola...Ch. 11.1 - Label the focus, directrix, and vertex on the...Ch. 11.1 - Graphs of Parabolas Match the equation with the...Ch. 11.1 - Prob. 6E Ch. 11.1 - Graphs of Parabolas Match the equation with the...Ch. 11.1 - Prob. 8E Ch. 11.1 - Graphs of Parabolas Match the equation with the...Ch. 11.1 - Graphs of Parabolas Match the equation with the...

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The center, vertices, foci and asymptotes of the hyperbola ( y + 2 ) 2 36 − x 2 64 = 1 .

Solution Summary: The author explains the center, vertices, foci and asymptotes of the hyperbola.

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To find: The center, vertices, foci and asymptotes of the hyperbola (y+2)236−x264=1 .

To sketch: The graph of the hyperbola (y+2)236−x264=1 .

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Chapter 11 Solutions